On algebraically closed field

Question

Prove that $K$ is a algebraically closed field iff there are not exist algebraic extensions over $K$ of degree $>1$

Can anyone tell me a hint to solve the problem?

Answer 1

We know (hopefully) that if $\;f(x)\in\Bbb K[x]\;$ is irreducible and has degree greater than zero, then $\;\Bbb F:=\Bbb K[x]\langle f(x)\rangle\;$ is an extension field of $\;\Bbb K\;$ with a root of $\;f(x)\;$ in it, namely $\;\alpha:=x+\langle f(x)\rangle\;$ , and the extension's degree is $\;\deg f(x)\;$.

Suppose now that $\;\Bbb K[x]\;$ is algebraic closed and let $\;f(x)\in\Bbb K[x]\;$ be irreducible of degree more than zero. But then $\;f(x)\;$ has a root in $\;\Bbb K[x]/\langle f(x)\rangle\cong \Bbb K[x]\;$ , which means $\;\deg f(x)=1\;$...

I leave it to you the other direction, which is more boring than the above one.